RESULTS: Erdős #126, the p-adic block-pruning arm (support run d176e366)

Hunt, not a result. Nothing here is evidence for or against RH (docs/08).

Headline, and it contradicts the brief. The target the brief names, "a block of $r$ primes can be removed while retaining more than $2^{-r+o(r)}$ of $A$", is not progress even if proved. Retention $2^{-r+o(r)}$ is exactly the Erdős–Turán bound restated (§4), and the block programme is not a proxy for Erdős #126 but a reformulation of it, exact to within a factor of 2 (§5). So the arm is well aimed and the threshold in the brief is set one notch too low: only $c^{-r}$ with $c<2$ buys a smaller base, and only $e^{-o(r)}$ settles the problem.

What is argued here (ordinary arguments, unreviewed; the run's draft said "rigorous", see the grading note below): the pruning problem is an independent-set problem (§3, Lemma 1), the joint block lemma holds at exactly $2^{-r}$ (§4, Lemma 2), that constant cannot be improved by any argument reading only residue data (§6, Theorem 3, with the extremal configurations characterised), and the whole programme is equivalent to the problem (§5, Theorem 4). What is not settled: no block lemma with base $<2$, and no unconditional obstruction. §8 says what it would take.

One correction to hunts/r_186989: its loose thread 3 is false (§2).

Landing note (2026-10-21). Landed on main as Hunt #94 from a local-only branch, support/d176e366, that was never pushed; PR #144 counted this arm among the seven that "produced no branch at all". The run's files are as it wrote them apart from punctuation and the dated notes marked as added on landing.

Grading (added on landing, 2026-10-21). Lemmas 1 and 2, the endgame lemma, and Theorems 3, 3′ and 4 are this lab's ordinary arguments, unreviewed and pending external verification; none is kernel-checked, and the run's labels are names, not grades. Theorem 3's formula matching exhaustive branch and bound four times out of four (§6) and Lemma 1's zero mismatches on 200 random instances (§3) are exact finite checks on blocks of at most three primes, not proofs. The §7 retention tables and the cross-prime correlation are measurements on witnesses with $|A| \le 11$, a direction indicator only, as the run says. In the terms of ALIGNMENT.md §5 the disposition is a restricted-class obstruction: Theorem 3 rules out beating base 2 for arguments that read only the residues of $A$ modulo $\prod_{p\in T}p$ (residue-only pruning), and says nothing about arguments that use the smoothness of the sums; the attempt below base 2 is unresolved. probe.py was re-run on landing and reproduced results.json byte for byte; no test in tests/ pins these numbers.

Overlap with arms already on main. That residue data cannot beat base 2 was reached on main by two other routes: Hunt #92 (support_517b887f), whose CRT configuration has $2^s$ points and largest jointly clean subset 1, and Hunt #101 (support_7ddfee4b), whose residue relaxation (the negation pairings of $k-1$ odd primes plus a triangle-free graph) has optimum exactly $2^k$. That relaxation is this arm's covering reformulation (Loose threads), there solved exactly. Theorem 4's verdict, that the programme restates the problem, is the kind of verdict #92 reaches for the 1934 architecture ($c(A)\ge|A|/\varphi(k)$ is equivalent to $g(k)\le 2\varphi(k)$), by a different quantity. Lemma 2 with $T=S\setminus\{2\}$ (§4) gives the bound $2^k$ of #101's primitive descent and of Hunt #98 (support_d0f0aefb Theorem 2, landed with this arm), here for the elements coprime to the odd primes of $S$. The refutation of r_186989's loose thread 3 (§2) was on main before this arm landed (Hunts #93, #95, #101, #103). This arm's own contributions are Lemma 1's independent-set form, Lemma 2 as a one-shot joint argument, Theorem 3's exact optimum on a full residue system with its check, Theorem 3′'s equality conditions, Theorem 4's two-sided equivalence $g(k)\le 2/\rho_{\min}(k)\le 2g(k)$, and the false sign-orbit claim kept with its counterexample (§6).

Reproduce with .venv/bin/python hunts/support_d176e366/probe.py (stdlib only, about a second on landing, writes results.json).

1. Setting and notation

$S = \{p_1,\dots,p_k\}$ a set of primes. $A$ a finite set of distinct positive integers is admissible for $S$ when every $a+b$ with $a\neq b$ in $A$ has no prime factor outside $S$. $g(k) = \max |A|$ over all such $S$ and $A$. Erdős #126 asks for $\log g(k) = o(k)$; Erdős–Turán (1934) give $g(k) < 3\cdot 2^{k-1}$.

For $T \subseteq S$ write

$$\varphi(A,T) \;=\; \max\{\,|A'| : A' \subseteq A,\ \gcd(a+b, \textstyle\prod_{p\in T} p) = 1 \ \ \forall a\neq b \in A'\,\}.$$

This is the quantity of the pruning programme: if $A'$ attains it then $A'$ is admissible for $S \setminus T$, so a lower bound on $\varphi$ that iterates is a bound on $g$. Write $\rho(A,T) = \varphi(A,T)/|A|$ for the retention ratio and $r = |T|$.

Throughout, $A^{\circ}T = \{a \in A : \gcd(a, \prod{p \in T} p) = 1\}$.

2. Audit of hunts/r_186989

Re-verified from scratch by trial division (check_witness_audit): all seven witnesses in that hunt's table are admissible, with the stated $S$, and each row's pairwise-sum prime support is contained in $S$. Its arm-3 lemma (no three positive integers have all three pairwise sums powers of $2$) is re-proved independently in §5 below and re-checked numerically over all exponent triples $x,y,z < 60$: no triples. An independent exhaustive re-run inside $[1,120]$ reproduces $g_N(k) = 2,4,5,6,8$ for $k=1..5$ with the same witnesses, which confirms that hunt's observation that the optima live in the small integers. (Annotated on landing, 2026-10-21: that observation did not survive on main. Admissible sets are closed under $S$-smooth dilation, so a wider box only adds larger copies of the same sets (Hunts #103 and #105), and a primitive extremal 5-set for $S=\{2,3,5\}$ reaches 475 (Hunt #107, $\{5,11,25,245,475\}$). What the re-run above confirms is the table rows and their witnesses, not where the optima live.)

One loose thread there is false. Its thread 3 proposes to "prove $|A| \le p-1$ when $p \notin S$ is small, by pigeonhole on residues mod $p$ against the pairing $r \leftrightarrow -r$." The pairing argument constrains which residue classes $A$ may use; it does not constrain how many elements sit in one class. For odd $q$ and $c \not\equiv 0$, two elements both $\equiv c$ have sum $\equiv 2c \not\equiv 0$, so a class may be occupied without limit. $A = \{1, 1+q, 1+2q, \dots\}$ has every pair sum $\equiv 2 \pmod q$ and is of unbounded size with $q \nmid a+b$ throughout (check_outside_prime_claim, sizes $3q$ for $q = 3,5,7,11,13$, verified). The one class where the pigeonhole does bite is $c \equiv -c$, i.e. $c=0$ for odd $q$ and both classes for $q=2$, which is why $2 \notin S$ forces $|A| \le 2$ and no odd $q$ does. That parity statement, which the same hunt reports separately, is the correct residual of the thread.

3. Lemma 1: the pruning problem is an independent-set problem

Lemma 1. Fix $T$ and let $G(A,T)$ be the graph on $A$ with $a \sim b \iff p \mid a+b$ for some $p \in T$. Then $\varphi(A,T) = \alpha(G(A,T))$, the independence number, and $G(A,T) = \bigcup_{p \in T} G_p$ where $G_p$ is the antipodal graph mod $p$: the disjoint union of the complete bipartite graphs between residue class $c$ and residue class $-c$ (for $c \not\equiv -c$), together with a clique on the class $c \equiv -c$ (i.e. $c=0$ for odd $p$; $c \in \{0,1\}$ for $p=2$).

Proof. Immediate from $\gcd(a+b, \prod T) = 1 \iff$ no $p \in T$ divides $a+b \iff a \not\equiv -b \pmod p$ for all $p \in T$. $\square$

Two consequences worth stating because they are what makes the arm's language ("valuation vectors", "ultrametric structure") not buy anything extra.

• Valuations beyond the first digit are invisible to this problem. Whether $p \mid a+b$ depends only on $a, b \bmod p$; $v_p(a)$ enters only through the indicator $v_p(a) = 0$. So the natural p-adic object here is the residue vector in $\prod_{p \in T} \mathbb{Z}/p$, not the valuation vector, and ultrametric refinement adds nothing until one asks for fixed $v_p(a+b)$ rather than $v_p(a+b) = 0$, which is §8's door 3.
• The clique on the zero class is the whole difficulty. If all of $A$ is divisible by $p$ then $G_p = K_{|A|}$ and one prime covers everything. That is not a counterexample (dividing $A$ by $p$ returns the same problem), but it is exactly why the covering bound of §7 does not close.

Checked (check_factorisation_lemma, 200 random instances, blocks $\{3,5\},\{3,7\},\{5,7\},\{3,5,7\},\{3,11\}$): the brute-force optimum equals the optimum over products of half-systems with the zero classes handled by hand. 0 mismatches.

4. Lemma 2: the joint block lemma, at exactly $2^{-r}$

Lemma 2. For any $A$ and any block $T$ of $r$ odd primes, $\varphi(A,T) \ \ge\ 2^{-r}\,|A^{\circ}_T|$.

Proof. For each $p \in T$ the $p-1$ nonzero classes split into $(p-1)/2$ antipodal pairs $\{c,-c\}$. Choose, independently for each $p$ and each such pair, one of the two classes uniformly at random; let $H_p$ be the union of the chosen classes, a random "half-system". These $\sum_{p} (p-1)/2$ coin flips are independent. For $a \in A^{\circ}_T$, $\Pr[a \bmod p \in H_p] = 1/2$ for each $p$, and the events over distinct $p$ depend on disjoint coin sets, hence are independent: $\Pr[a \in \prod_p H_p] = 2^{-r}$. So $\mathbb{E}\,|A \cap \prod_p H_p| = 2^{-r}|A^{\circ}_T|$ and some choice attains it. Any such set is independent in $G(A,T)$ by Lemma 1, since $c, -c \in H_p$ never both hold. $\square$

This is genuinely joint (one shot over all $r$ primes, no sequential conditioning), and it lands on exactly the same constant as pruning one prime at a time. That is not a coincidence; §6 shows it cannot be beaten from residue data.

Applied with $T = S \setminus \{2\}$ and combined with Theorem 4's $\varphi \le 2$: $|A^{\circ}_{S\setminus\{2\}}| \le 2^{k}$. That is weaker than Erdős–Turán, which also controls the elements divisible by a prime of $S$; the random half-system argument never captures those (they are in no $H_p$), and closing that gap is the zero-class clique of Lemma 1 again. We did not close it.

5. Theorem 4: the block programme is the problem

Endgame lemma. If $A'$ is admissible for $S$ and every pair sum of $A'$ is coprime to $S \setminus \{2\}$, then $|A'| \le 2$.

Proof. Every pair sum is $S$-smooth and coprime to every odd prime of $S$, hence a power of $2$ (if $2 \notin S$, a pair sum would be $1$, impossible, so $|A'| \le 1$). Suppose $a<b<c$ in $A'$ with $a+b = 2^x$, $a+c = 2^y$, $b+c = 2^z$. Then $a+b < a+c < b+c$ gives $x<y<z$, so $2a = 2^x + 2^y - 2^z \le 2^x + 2^y - 2^{y+1} = 2^x - 2^y < 0$, impossible. $\square$

(Independently re-derived; hunts/r_186989 has the same argument. Numerically re-checked over all $x,y,z<60$: no triples.)

Theorem 4 (equivalence). Let $\rho_{\min}(k) = \min \{\varphi(A, S\setminus\{2\})/|A|\}$ over all $S$ with $|S| = k$ and all $A$ admissible for $S$. Then

$$g(k) \;\le\; \frac{2}{\rho_{\min}(k)} \;\le\; 2\,g(k),$$

and therefore $\log g(k) = o(k) \iff \rho_{\min}(k) = e^{-o(k)}$.

Proof. Left: for any admissible $A$, $|A| = \varphi/\rho \le 2/\rho(A) \le 2/\rho_{\min}(k)$ by the endgame lemma. Right: $\varphi \ge 1$ always and $|A| \le g(k)$, so $\rho(A) \ge 1/g(k)$ for every admissible $A$, hence $\rho_{\min}(k) \ge 1/g(k)$. $\square$

What this settles about the brief. Retention $2^{-r+o(r)}$ at $r = k-1$ gives $g(k) \le 2^{k+o(k)}$: the Erdős–Turán bound, with a worse $o(k)$. The block lemma is worth something only strictly below base $2$. Nothing in the brief's framing is wrong; its threshold is.

6. Theorem 3: the obstruction to residue-only pruning

Theorem 3. Let $T$ be a block of $r$ odd primes, $M = \prod_{p\in T} p$, and let $A$ be a full residue system mod $M$ (e.g. $A = \{1,\dots,M\}$). Then

$$\varphi(A,T) \;=\; \prod_{p \in T}\frac{p-1}{2} \;+\; r \;=\; |A|\prod_{p\in T}\frac{p-1}{2p} \;+\; r,$$

so the retention ratio is $\prod_{p}\frac{p-1}{2p} + r/M$, which is strictly below $2^{-r}$ as soon as $M$ is large.

Proof. Upper: by Lemma 1 an independent set uses, mod each $p$, at most one class from each antipodal pair, plus at most one element in class $0$; the elements with a zero coordinate number at most $r$ (one per prime), and the rest lie in $\prod_p H_p$ for some half-systems, of which there are $\prod_p (p-1)/2$ residues mod $M$, one element each. Lower: take those $\prod_p (p-1)/2$ elements, and for each $p \in T$ add by CRT an element $\equiv 0 \bmod p$ and in $H_q$ for every $q \neq p$; all new pairs are checked coordinatewise. $\square$

Verified against exhaustive branch-and-bound for $T = \{3,5\}, \{3,7\}, \{5,7\}, \{3,5,7\}$ (check_tightness): predicted $4, 5, 8, 9$, brute force $4, 5, 8, 9$. Exact match, four for four.

(The one row above $2^{-r}$ is $15 \cdot \tfrac{1}{3}\cdot\tfrac{2}{5} = 2$ plus the additive $r=2$; the main term is below $2^{-r}$ in every row and the additive term is $O(r/|A|)$.)

Theorem 3' (when Lemma 2 is tight). Suppose every element of $A$ is coprime to $M$, and give $\prod_{p\in T}(\mathbb{Z}/p)^\times$ the coordinatewise sign action of $\{\pm 1\}^T$, whose orbits all have size $2^r$. Write $N(\varepsilon)$ for the number of elements of $A$ captured by the half-system $\varepsilon$. Then:

1. $\mathbb{E}\varepsilon N = |A|2^{-r}$, so $\varphi(A,T) = \max\varepsilon N \ge |A|2^{-r}$ (Lemma 2 again), with equality iff $N$ is constant;
2. if the residue image of $A$ is a union of full sign-orbits, each occupied with the same multiplicity, then $N \equiv |A|2^{-r}$ and Lemma 2 is tight;
3. equality forces $|A| \ge 2^r$, since $\varphi \ge 1$.

Proof. (1) is the computation of Lemma 2. (2): a half-system meets each full orbit in exactly one point. (3) is immediate. $\square$

What is not true, and cost us a draft. $\varphi(A,T)$ is not the number of sign-orbits $A$ meets, even though each orbit is captured by one bit per prime. Distinct orbits can share a bit: mod $(3,5)$, the elements $\equiv(1,1)$ and $\equiv(2,2)$ lie in different orbits but both need the sign at the pair $\{1,2\}$ mod $3$, and they need it opposite. Two orbits met, $\varphi = 1$. Independence across primes is real (Lemma 2); independence across orbits is not. This is why §3's independent-set formulation, and not any closed form, is the right object.

What Theorem 3 and 3' rule out. Any argument that lower-bounds $\varphi(A,T)$ using only the reduction of $A$ modulo $\prod_{p\in T}p$ is capped at $\prod_p \frac{p-1}{2p} + r/|A| < 2^{-r} + r/|A|$, because a residue configuration attaining that cap exists and the argument would have to be valid for it. So no coordinatewise-or-joint pruning scheme that ignores the smoothness hypothesis can beat base $2$, and by Theorem 4 no such scheme can improve on Erdős–Turán. Any improvement must prove that an admissible $A$ cannot look like the extremal configuration.

And that is not vacuous, in a way worth naming. Tightness in Theorem 3' forces $|A| \ge 2^r$ and a capture count that does not vary with the half-system. An admissible $A$ in the regime of interest is far smaller than $\prod_{p\in T} p$, so the slack the programme needs is real and it is located precisely at: how far is the residue image of an admissible $A$ from being closed under coordinatewise negation? That is the door.

7. The measured signal, and why it is only half a signal

For each verified witness we computed $\varphi(A,T)$ exactly for every block $T \subseteq S \setminus \{2\}$ (check_full_block_pruning).

Retention ratios, $k=7$, $|A|=11$:

The $r=1$ ratio sits slightly below $1/2$ (consistent with $(p-1)/2p$), yet the $r=2$ ratio is $0.364$ where independence would predict $\approx 0.45^2 \approx 0.20$. Survival across primes is strongly positively correlated in every measured witness, and the gap widens with $r$: at $r=6$ the ratio beats $2^{-r}$ by a factor of $11.6$. Same picture at $k=5,6$.

The honest caveat, and it is severe: at the full block $r = k-1$ the ratio is $2/|A|$ by Theorem 4, so that column is the conjecture restated at small $k$ and is not independent evidence. The intermediate columns ($r=2,3,4$) are not forced by Theorem 4 and do carry information, but $|A| \le 11$, and finite data alone is not progress. We report it as a direction indicator only.

check_full_block_pruning also confirms $\varphi(A, S\setminus\{2\}) = 2$ for all seven witnesses, as the endgame lemma requires.

8. What we could not settle, and what it would take

• No block lemma with base $< 2$. Lemma 2 is joint and lands exactly on $2^{-r}$; Theorem 3 says residue data cannot do better. Not settled.
• No unconditional obstruction. Theorem 3 is an obstruction to residue-only arguments, not to all arguments, precisely because the extremal configuration it exhibits is not realisable by an admissible $A$ once $|A| < 2^r$. A stronger obstruction would need extremal configurations that are themselves admissible, and §7's measurements say those do not exist at small $k$.
• The zero-class gap in Lemma 2. The random half-system never captures an element divisible by a prime of $T$, so Lemma 2 bounds $|A^\circ_T|$ and not $|A|$. Erdős–Turán handle this; we did not.

The doors

This hunt measured a ceiling (the $2^{-r}$ retention wall), so it owes the list.

1. Active constraints at the optimum.

2. Frozen-constant inventory.

3. Information class. Doors 2 and 3 stay inside what the current family reads (residues mod $\prod T$) and are therefore under Theorem 3's ceiling: they can change the $e^{o(r)}$ and never the base. Door 1 (fixed valuations / the p-adic tree) and the covering reformulation below require reading more: deeper p-adic digits, or the size constraint $a+b \le 2\max A$ together with $\sum_p v_p(a+b)\log p = \log(a+b)$, which is the only place the smoothness hypothesis (as opposed to the divisibility hypothesis) has entered nothing so far. That is where the base-$2$ wall actually sits.

Loose threads

• The covering reformulation. By Lemma 1 plus the endgame lemma, $A$ admissible for $S = \{2\} \cup T$ forces $K_{|A|} = G_2' \cup \bigcup_{p\in T} G_p$ where $G_2'$ is the graph of pairs whose sum is a power of $2$, which is triangle-free by the endgame lemma. Off the zero classes each $G_p$ is a disjoint union of complete bipartite graphs, so Hansel's bipartite-covering bound applies and reproduces $|A| \le 2^{r}$ up to the $G_2'$ handling. Why it might matter: bipartite-covering bounds have refinements (Radhakrishnan–Srinivasan) that are sensitive to exactly the structure $G_p$ has, and the problem becomes finite combinatorics rather than number theory. First step: bound $\alpha(G_2')$ or the covering cost of a triangle-free graph well enough to avoid the $\sqrt{|A|}$ loss of the naive independent-set route, then run Hansel on the rest.
• The smoothness hypothesis has not been used at all. Everything above uses only "$a+b$ is divisible by primes of $S$", never "$a+b$ is exhausted by primes of $S$", except in the endgame lemma. Why it might matter: it is the entire remaining slack, and the size constraint $\sum_p v_p(a+b)\log p = \log(a+b) \le \log(2\max A)$ is a genuine budget across primes that no pruning argument here has spent. First step: bound the number of admissible $A$ with $\max A \le X$ via $S$-unit equation counts, which is the door hunts/r_186989 already ranked first, and which this hunt did not touch.
• Weighted half-systems against real witnesses. For each measured witness, compare Lemma 2's uniform choice against a frequency-weighted one and see how much of §7's factor $11.6$ a proof technique can recover. Why it might matter: it separates "the correlation is real" from "the correlation is reachable by an argument". First step: a 20-line addition to probe.py.